Talk:Operational amplifier/Archive 1

Common configurations list
I added some links which describe the basic opamp configurations. I am thinking of including them in the article itself, along with diagrams, equations for gain, input impedance, etc. but i am not sure if that will be too in depth for a wikipedia article. I am not talking about a huge article with detailed derivations of each equation, just a listing of configurations and naked equations to be used as a reference. There are a lot of sites that have this, but none have all of the information for all of the main configurations in one place, and most have derivations, or are school sites with homework problems and not the actual solutions. Should I undertake this or just leave the links? - Omegatron 19:29, Feb 18, 2004 (UTC)

Perhaps I will make an article Standard operational amplifier configurations and link it from here. - Omegatron 17:03, Feb 27, 2004 (UTC)

I would like some opinions, please. I am making a list of the common configurations, to be used as a reference, and I don't know if I should put it here or make a new article for it, since it is somewhat detailed and long. It is on my user page for now, incomplete. I guess I will stick it in this article if no one says anything. - Omegatron 14:40, Mar 23, 2004 (UTC)

As of July 20, I think this is an extremly well written article. I would like to see the standard Op Amp diagram used in the Schmidt Trigger article, where the power supply inputs are left out, and it is never made clear that there are are two supplies of opposite polarity and equal magnitude.--Don 20:39, 20 July 2006 (UTC)


 * I think it would be good to eventually have a separate article on each common configuration, with a very basic description in the main article. Many of these op-amp circuits are widely enough used to be the basis for encyclopedic, albeit somewhat technical, articles.  CyborgTosser


 * I'm an interested know-nothing in this field (though I am a software engineer, so I am capable of some technical understanding) and I have to say this article kind of stunned me with foreign language and detail right off the bat. I found the "common applications" page and then found this one that simply said "Compares two voltages and outputs one of two states depending on which is greater" aside a diagram that shows two separate inputs and an output. Is that what an op amp does?  If so, perhaps that phrase would be a candidate for inclusion in the opening paragraph?  It's very concise and lays some groundwork for a novice that (for me anyway) was very helpful SVDasein


 * Well, theoretically, one of the things WP is not is a textbook, so I didn't want to go too in depth about it, and just make a list that could serve as a reference. (When i go to look for this stuff online i invariably find homework assignments that just want me to re-figure it out myself.)  But I guess we could for the very common ones. - Omegatron

Probably no one cares, but there are two different versions of the common configurations section, different only in formatting, at User:Omegatron/opamps. Which is better? - Omegatron 18:09, Jun 14, 2004 (UTC)

New graphics/notation
I think it would look better if the voltages were labeled with subscripts and +/- as follows:

Original:



vo = (-K * vo) + vp

vo = 1/(1+K) * vp

New:



Vout = (-K * Vout) + V+

Vout = 1/(1+K) * V+

Does anyone like the original graphic and voltage notation better? Omegatron 18:57, Feb 27, 2004 (UTC)


 * I ended up just being bold and changing this. - Omegatron 14:40, Mar 23, 2004 (UTC)


 * I think the new version is much improved. Very good for a web page, really. --blades 23:40, Mar 24, 2004 (UTC)

Op-amp internals
A really elementary question - how would one build a op amp out of transistors, resistors, etc? All the descriptive material seems kind of black-boxy to me. Stan 19:41, 22 Mar 2004 (UTC)


 * There are a lot of different designs. We could cover the basic stages, at least.  I believe it is a differential amp followed by a high gain amp followed by a unity gain power amp.  I could definitely be wrong.  I believe that the individual stages can be of several different types, FET vs BJT inputs, for instance.  Also there are many additions in modern op-amps to make them easier to use, like short circuit protection.  Here is the first schematic I found on google, to illustrate the complexity:


 * http://www.bausch-gall.de/images/prodss03.gif


 * - Omegatron 22:28, Mar 22, 2004 (UTC)


 * Ooh, my EE knowledge paging back in, makes me feel twenty years younger... So maybe a block diagram of the amps, plus one transistor-level schematic as an example of why black-boxing it is good. :-) (Also helps explain why they were less popular before the days of ICs...) Stan 22:38, 22 Mar 2004 (UTC)


 * Yes. Block diagram is good.  And yes, they are complicated.  Originally they were made with tubes!  I'm glad I'm such a young'n, and  never had to deal with such things. :-)


 * Hmm. Can we use a transistor level diagram from a datasheet or is it copyrighted?  http://www.national.com/ds/LM/LM741.pdf  (page 4)  741 is pretty standard and relatively simple.  Maybe this one? - Omegatron 22:52, Mar 22, 2004 (UTC)


 * I would think they'd be copyrighted, and when WP is world-famous, National will notice... It would be OK to make one's own schematic of the 741 using their datasheet as a source; a bonus is that you can make it large and thumbnail it in the article, so we end up with something more readable than the usual muddy datasheet blob. Alternatively, one could always write to a manufacturer and ask them to donate a schematic; free publicity for them. It might be fun to write to several with a single message so that each can see the cc's to the other, get a little competitiveness going... Stan 13:09, 24 Mar 2004 (UTC)


 * The thing is that these amplifiers are standards. Every company makes a 741 and probably most of them are exactly the same.  But yeah, it would be really easy to make a klunky schematic version and avoid any trouble.  I will do it, but it will take me a few days to get around to it. - Omegatron 17:09, Mar 24, 2004 (UTC)


 * Done.




 * Anything else I should add? We should outline the different sections in colored dashed lines and labels.  Someone please double check it.  Most probable mistakes are NPN/PNP and wiring mistakes (especially crossing wires vs 4 way junction), but I think it is right. - Omegatron 14:36, Mar 25, 2004 (UTC)


 * Cool! Functional blocks shown with solid rectangles of light colors would make it perfect. Stan 17:11, 25 Mar 2004 (UTC)

Note to self: http://www.national.com/appinfo/amps/files/Opamp_Trivia_WEBCAST_FINAL.pdf page 47. ;-)

Or I could color code the circuitry itself, similar to their highlighting specific parts in dark black. Would this be easier to read? Do we have to worry about color blindness (we would have to with dashed outlines too) (i think we would just have to avoid certain colors next to each other?)

- Omegatron 20:06, Mar 25, 2004 (UTC)

Alright. I did what I could:


 * Oooo look! I just found the resistor acting as a current source (39k). Not perfect, but hey, if they use it in an integarated circuit as the main current source it must be good enough to convince anyone. Light current 00:40, 28 August 2005 (UTC)
 * I think that's a stretch. The 39k resistor sets the reference current but not by sourcing or sinking current - the power supplies do that - but by limiting the current in the same way that any series resistance does according to Ohm's law. I suppose what you are getting at is the Thevenin / Norton duality - a voltage source with a series resistance is equivalent to a current source with a parallel resistance. Alfred Centauri 16:05, 30 August 2005 (UTC)
 * Sorry, I was just harking back to and old discussion between me and Omegatron about what can be considered to be a current source. Light current 17:45, 30 August 2005 (UTC)

I don't know everything. Specifically the magenta section. Is it part of the lowpass filter? And what does that extra transistor on the output do? I can alter the image if some of the sections should be a little different. Please add what you know to the internals section. - Omegatron 17:06, Jul 17, 2004 (UTC)

The Magenta Section
I think that in both cases we have constant-current controllers.

Take the transistor in the output section that has a 25&Omega; resistor between its legs: If the bias current through the 25&Omega; were to increase for some reason, its base-emitter voltage would rise above, say 0.6V, the transistor would turn on more and so it would draw more collector current - out of the main output transistor's base. This would turn that off a bit and thereby reduce the bias current again. The three transistors in the magenta section are in the same configuration except that two of them seem to form a kind of 'toned-down' darlington pair (toned down by the 50k, which reduces their combined forward gain slightly). The third one 'saps away' base current out of the first of the pair until a happy 0.6-0.7V appears across the 50&Omega; resistor. This sets the bias current in the green driver stage.Any help?Nigelj 15:43, 28 Aug 2004 (UTC)


 * Magenta section looks like Class A gain stage (darlington) with current limiting provided by TR that has 50R from its emitter to its base. Cyan section is the ouput stage with +ve current limiting on the top transistor. Light current 18:19, 27 August 2005 (UTC)
 * That sounds right to me - the current limiting in the driver stage can only protect the PNP output transistor - positive going drive current is sourced by the current mirror thus the NPN output transistor needs a separate protection circuit. Alfred Centauri 16:15, 30 August 2005 (UTC)

''

The cyan section
I still dont fully understand the 50R in the output stage. Any body know its purpose (apart from simple current limiting for negative outputs?? Light current 19:31, 30 August 2005 (UTC)

Offset Null Adjustment
''Having a bit of trouble using the offset null on the 741 amp. Is it possible for someone to walk me through it? Thanks''

It's important to use the Offset Null adjustment correctly in order to get the best out of any op-amp that has one. Setting it up correctly will optimise two aspects of performance:

Offset Drift
The dc output voltage of a real op amp for any given input conditions will drift with both temperature and ageing. Correctly setting the offset null adjustment will help to minimise this in both cases.

Common-Mode Rejection Ratio (CMRR)
The dc output voltage of an ideal op-amp would be zero when the two input voltages were at the same voltage, no matter what that voltage was. In practice, there will be some change in output as the voltage of the two inputs is varied, even though they remain tied together. The ability of the amp to reject these 'common-mode' inputs will also be optimised by correctly nulling the input offset

Setting up the 741
(What follows applies to the standard &mu;A741 of days of yore. Look up the manufacturer's data sheet for whatever amp you're using and follow the correct recommendations!) Connect a 10k&Omega; linear pot between pins 1 and 5, and connect its wiper to the negative supply at pin 4. Connect both inputs (pins 2 and 3) to 0V. Note, these are the actual op-amp inputs NOT the inputs you are going to expose to the outside world, the other side of whatever input resistors. Adjust the pot until there is no voltage between pin 6 and 0V.

Important Note
Regarding the note in the the previous paragraph, it is often necessary to design in another 'zero' adjustment, which may inject a small +/- voltage into a summing junction at one of the inputs. It is important then that the 10k input offset pot is adjusted as described above to null the op-amp's actual input offset, so as to get the performance benefits described earlier. The zero adjustment is then used to create the desired behaviour of the overall circuit, without touching the input offset adjustment again.

Nigelj 16:12, 24 Oct 2004 (UTC)

Vs+/Vs- are reversed?
The voltage supply rails are mislabeled aren't they? The top one should be Vs+, the bottom Vs-.


 * Oops. Good eye.  Fixed. - Omegatron 16:33, Jan 31, 2005 (UTC)
 * hi i want to know why op-amp give -15 and +15 volt supply.

Differencing input impedance
Who knows the differential input impedance of the differencing amp?



I've been told that my original calculation was wrong by someone who is usually right, so I commented it out for now. I am skeptical and think I was right, though. - Omegatron 06:32, Mar 18, 2005 (UTC)


 * These comments apply to single ended output version of the circuit:


 * For a floating voltage source connected between the input terminals (a microphone for example), the resistance seen by the source is simply the sum R1 + R2. User:Alfred Centauri 03.18 19 Apr 2005


 * Excellent. That's what I had previously but a senior engineer hesitantly disagreed.  I think they were just confused... - Omegatron 18:12, Apr 16, 2005 (UTC)


 * What is interesting about this configuration is that the input terminals will generally have a common mode voltage present that is a function of the output voltage. In fact, for the case R1 = R2, R3 = R4, the input common mode voltage is given by Vout / 2.

The presence of this "Vout induced" common mode voltage on the input terminals and thus the cable connecting the source to the amplifier is not desirable. I have recently shown that it is possible to find values for the four resistors such that for a specified input resistance and difference gain, the common mode gain is zero and the input common mode voltage due to Vout is zero.

For example, a balanced 600 Ohm microphone pre-amp with a voltage gain of 20 requires the following 1% resistor values:


 * R1 = 590, R2 = 14.3, R3 = 11.8k, R4 = 287


 * Compare these with the 'textbook' values of:


 * R1 = 300, R2 = 300, R3 = 6000, R4 = 6000


 * For both sets of resistor values, the pre-amp will have a nominal 600 Ohm input resistance, a difference voltage gain of 20 and a common mode voltage gain of 0. However, the 'textbook' pre-amp will generate a common mode voltage at the input terminals that is 10 times the source voltage!User:Alfred Centauri 03.18 19 Apr 2005


 * Excellent info. Can you add it to that section of the article? - Omegatron


 * I'd be happy to!User:Alfred Centauri 03.18 19 Apr 2005


 * Hmm.. Actually I'm confused about where the common-mode voltage appears.  Can you describe the rest of the circuit?  Where is the common-mode voltage and what is it relative to? - Omegatron 18:16, Apr 16, 2005 (UTC)


 * The common mode voltage on the input terminals is defined as: (V1 + V2) / 2, the average of the two input voltages.  Here's a physical picture.  Connect a voltmeter between V1 and ground.  Connect a second voltmeter between V2 and ground.  Let's say that V1 reads 9.5V and V2 reads 10.5V.  The difference voltage is: (10.5V - 9.5V) = 1V.  The common mode voltage is: (10.5V + 9.5V) / 2 = 10V.  In words, it is the voltage common to both input terminals.User:Alfred Centauri 03.18 19 Apr 2005


 * I'm more confused now. Isn't that what's supposed to happen?  That's how the difference amplifier works.  There will always be a voltage on the opamp terminals because of the R2/R4 divider.  You're just minimizing it by making R4 >> R2? I know the properties of an amplifier depend on the common mode voltage, and the designer tries to minimize this, but does that apply in this case? - Omegatron 23:22, Apr 17, 2005 (UTC)


 * I think the confusion may be that you are referring to the opamp input terminals (V+, V-) which, as you know, must have the same voltage at all times (virtual short) when negative feedback is present. As you correctly point out, this is a pure common mode voltage.  However, in the comments above, I am referring to the common mode voltage at the input terminals of the circuit (V2, V1).  The common mode voltage gain (Acm) of the circuit (not the opamp) is defined as:  2(Vout) / (V1 + V2).  The CMRR of the opamp is a function of the internal circuitry.  For this circuit, the Acm depends on the external resistor values, not the CMRR of the opamp!  It's easy to see from your fundamental equation for Vout that Acm can be made the desirable value of zero if the following condition is met

(1) [(R3 + R1)R4 / (R4 + R2)R1] = R3 / R1

Not surprisingly, the condition for the amplified difference, R1 = R2, R3 = R4, also satisfies the constraint (1) for Acm = 0. Unfortunately, this arrangement generates a voltage (with respect to ground) of Vout / 2 on both the V1 and V2 terminals. Picture this: you connect a microphone to the V2 and V1 terminals of this amplifier circuit with the 'textbook' resistor values I gave above. Let's say that the microphone produces a 100mV AC voltage ACROSS V2 and V1. If you measure the voltage with an oscilloscope from V2 to ground, you will see the original microphone signal amplified by 10.5! If you connect a second oscilloscope probe to V1, you will see the original microphone signal amplifed by 9.5 and that the two voltages are in phase. If instead, you use the resistor values I gave, you would measure the microphone signal divided by two on both V2 and V1 and that this signal is truly differential. That is, the two voltages are out of phase - just what you would expect to see coming out of the microphone when not connected to the pre-amp.
 * Do you use circuit simulator software such as pSPICE or an equivalent? If so, I can send you the circuit files that I used to simulate the two circuits.User:Alfred Centauri 03.18 19 Apr 2005

Differencing amplifier with floating output?


These comments apply to the floating output version of the circuit:

It seems to me that the floating output version of the diff-amp is not, in general, a valid circuit. Connect a floating voltage source, Vin, between the input terminals. The open circuit output voltage, Vout(oc), must equal Vin. The reason is that there is no path for source current and so the voltage across every resistor in the circuit is 0V. With a load resistance connected between the output voltage terminals, the circuit cannot be solved unless Vin is exactly zero.

If each input terminal is connected to a voltage source referenced to ground, you do have a valid circuit that contains both negative and positive feedback. This means that with the appropriate choice of resistor values, the voltage gain becomes unbounded i.e., the amp becomes unstable.


 * Oops. I saw it on some website and thought "whoa!  a balanced output configuration!".  I guess it's too good to be true... - Omegatron 18:12, Apr 16, 2005 (UTC)


 * Here it is: http://webpages.ursinus.edu/lriley/ref/circuits/node5.html - Omegatron 14:17, Apr 18, 2005 (UTC)


 * I've taken a quick look at the web page above and have come to the conclusion that the end of R4 that appears to be floating in the diagram above is actually the implied common node (usually the power supply ground). To see this, look at his diagram for the voltage follower (figure 26) just above his diagram for the differential amplifier circuit in question.  Note that he shows Vin and Vout referenced to a floating node (the line isn't connected to anything)!


 * The only way to make any sense out of this diagram (and the others on the page) is to assume that the line representing that floating node is actually the power supply common (ground). Otherwise, there is no path for current through a load!  It is conventional to assume that the voltage at the output node of the opamp is referenced to the power supply common.  Thus, there should be a ground symbol connected to that floating node.  Further, take a look at the very first diagram (figure 22) on the page.  The author shows the negative terminal of the dependent voltage source as an external, floating connection.  Where is that connection in the rest of the circuit diagrams?  That terminal should be shown as an internal connection to ground, IMHO.  Then, it is clear that the path for load current is through the dependent voltage source in figure 22.


 * Alright. So I just invented it.  I've already removed it from the article. - Omegatron


 * In order to have a true balanced (differential) output, I believe that you must use two opamps with the load connected between the two output terminals. This is done all the time with stereo power amplifiers.  You drive the inputs of the stereo amp with equal but opposite voltages and then connect the speaker between the two outputs of the amplifier.


 * Yes, the bridged amplifier. For opamps to simulate a transformer-coupled floating output, we use the "cross-coupled" connection, right?   Like so: http://www.rane.com/n124fig5.gif
 * I just saw that diagram, assumed it to be true, and thought "Wow, a one-opamp version of the cross-coupled output!"
 * You know, you can sign your talk page posts by typing or ~ for name plus timestamp. - Omegatron 17:19, Apr 18, 2005 (UTC)


 * That 'cross-coupled' circuit was fun to analyze! I found that the overall voltage gain of the circuit is 2 but that interestingly, the voltage gains for Vpos and Vneg are not equal but opposite.  Vpos / Vin is 1.5 while Vneg / Vin is -0.5 (an inverting attenuator!).  This gain imbalance produces (you guessed it) a common mode OUTPUT voltage equal to Vin / 2.  To exactly simulate a transformer-coupled floating output, the common mode output voltage should be zero which requires the voltage gains for Vpos and Vneg to be equal and opposite.  You've got me thinking about this when I should be doing something else!  Nonetheless, I'm going to investigate this circuit to see if there is a resistor combination that fixes this problem.


 * Thanks for the tip! Alfred Centauri 19:48, 18 Apr 2005 (UTC)


 * I am now suspicious that the single-ended gain figures I gave above are an artifact of the circuit simulation software. When I first solved the equations for this circuit, I came up with (Vpos - Vneg) = 2 Vin.  This equation only specifies the voltage difference between Vpos and Vneg.  To specify Vpos or Vneg w.r.t. ground, we need another equation but for the life of me, I can't seem to get another independent one.  Although I haven't confirmed this, my belief is that the positive feedback exactly cancels the negative feedback so that we no longer have a unique solution.  The circuit simulator is giving the first solution it converged to.  Bottom line, it appears that all we can say is that if Vpos / Vin = Apos where 0 <= Apos <= 2, then Vneg / Vin = Aneg = -(2 - Apos).  I found some mention of this circuit on the web and the impression I got is that it is a very difficult circuit to balance and to keep balanced.  Doesn't surprise me a bit!  Alfred Centauri 22:57, 18 Apr 2005 (UTC)

Do you have to put a voltage source between ground and one of the outputs? - Omegatron 01:38, Apr 19, 2005 (UTC)


 * The interesting thing about this circuit is that the differential output voltage (Vpos - Vneg) is unaffected by putting a voltage source or a ground on one of these terminals. If you connect Vneg to ground, Vpos is simply 2 Vin.  If you connect a voltage source of K Volts to Vneg, Vpos is:  (2 Vin) + K.  Either of these connection 'forces' Apos to become 2.  Unfortunately, both of these connections defeats the intended purpose of the circuit - single-ended to balanced conversion. Alfred Centauri 02:18, 19 Apr 2005 (UTC)


 * I meant "do you have to provide a voltage reference for one of the output pins in order for the software to validate it?" — Omegatron 19:45, August 27, 2005 (UTC)
 * Oh! Do you mean for the simulator to converge?  Hmmm...  I think the simulator will converge to a solution out of the infinity of possible solutions without the help of the voltage source.  If you put a voltage source on one of the output pins, you force it to pick one solution. It's been awhile since I played with this circuit and my hard drive failed recently so I no longer have the circuit file to try.  I need to recreate it anyway so I'll let you know.  Alfred Centauri 01:09, 31 August 2005 (UTC)

Instrumentation amplifier


Is this right? Should I add any other labels? Is the Vin polarity correct? I've seen both online... I've also seen ones where R2 = R3. - Omegatron 02:02, Apr 19, 2005 (UTC)


 * The polarity of Vin is fine but this circuit is inverting: Vout = -Vin(1 + 2R1 / Rgain)(R3 / R2).   Setting R2 = R3 simplifies the gain formula to:  Vout / Vin = Adiff = -(1 + 2R1 / Rgain).  Clearly, if you define Vin with the opposite polarity, the gains become positive (non-inverting).  The nice thing about this circuit is that no input common mode voltage is generated by Vout.


 * I think your diagram looks great! I wouldn't change a thing. Alfred Centauri 02:44, 19 Apr 2005 (UTC)

Input stage Explanations
It would be very handy if all transistors in the 741 diagram had references, like Tr1, Tr2 etc. Then we could all talk more easily about the circuit. Are you able and willing to do this Omegatron? Light current 00:09, 28 August 2005 (UTC)

Im going to refer to the transistor reference numbers quoted here in future to save any confusion. Light current 20:01, 29 August 2005 (UTC)

Just noticed that National have got some errors on their transistor numbering!! eg the Vbe mult doesnt have a number and there are two Q15 s! No Q16 either. Tut tut! 12:45, 30 August 2005 (UTC)
 * Certainly I can add some numbers. I will use that numbering but which one should be Q16?  — Omegatron 15:54, August 30, 2005 (UTC)

I suggest the following changes to NatSemis digaram : Q17 becomes Q19, 'Vbe mult' transistor becomes Q16, Q15 becomes Q 17. Does that meet with everyones approval? Light current 16:25, 30 August 2005 (UTC)


 * Which 15 becomes 17? — Omegatron 01:08, August 31, 2005 (UTC)

Sorry, its the rightmost Q15 becomes Q17 (I forgot there were 2 of 'em) Light current 01:18, 31 August 2005 (UTC)


 * Yick. I think it's far too cluttered, and I only labeled the transistors, but it will work for now: Image:Opamptransistorlevelcoloredlabeled.png
 * Maybe this section will deserve its own article someday and it will get a better image. Maybe someday the open-source community will actually finish one of the myriad stale circuit-drawing softwares.  (Not likely.)  — Omegatron 01:37, August 31, 2005 (UTC)

No its not too bad ,O. Maybe if the font size was reduced it would look better. BTW I just noticed we dont have a Q18, but we do have a Q22 (Shoot!!) Light current 01:56, 31 August 2005 (UTC)

Input Stage Description
Is anyone 'in the know' able to offer a detailed explanation of the operation of the input stage. ie. what is the purpose of all the transistors in the 741 i/p stage?. I think the article could do with this detail. Light current 12:09, 28 August 2005 (UTC)

Im not convinced of the statement that 'the input stage is loaded by a current mirror'. If so, which current miror acts as the load?. Also, where do people think the output of the input stage is? Again, how does the cct provide differential to single ended conversion? A lot more explanation of this cct is needed. Light current 22:52, 27 August 2005 (UTC)

The Bottom 2 transistors (Q5 & Q6) in input stage have the 'same' collector current. (because their bases are tied together and they have equal emitter resistors). This should give a clue to i/p stage operation. This current seems to be determined by the NPN emitter follower transistor (Q7) which in turn has its base connected to the collector of the lower left transistor. If the voltage at this latter collector rises due to an increase of voltage on the non inverting input, then the current increases in the lower right transistor and causes the voltage at the output of this stage to decrease. By contrast, if the voltage on the inverting input rises, the upper right transistor turns on more and tends to increase the stage output voltage. Hence there appears to be a conversion from diff mode to single ended here. Does anyone have any comments on this argument so far?? Light current 15:52, 29 August 2005 (UTC)

The lower current mirror (Q10 & Q11)output seems to provide part of the bias current for the central two transistors in the input stage. The upper left current mirror may be something to do with the common mode handling as it also has an influence on the bias of the central pair of transistors. If anyone knows more on this i/p stage, could they please comment on the above? Light current 16:20, 29 August 2005 (UTC)


 * I'm not sure how the transistors are labled so I'll go from top to bottom...
 * The top two transistors form a current mirror that samples the sum of the collector currents for the emitter follower input transistors below. These emitter follower transistors drive the common base amplifiers below them.  These four transistors together form a diff amp.  The bottom three transistors form a classic 'base-current compensated current mirror' that acts as an active load for the diff amp and that converts the differential signal currents into a single-ended output voltage at the collector of the transistor that is connected to the driver stage.  The DC bias for this stage is set by the base voltage for the common base amplifiers.  Because of the feedback from the current mirror at the top of the input stage, this voltage will be whatever it needs to be such that the sum of the collector currents for the input transistors equals the (constant) current in the current mirror just to the left of the driver stage. Alfred Centauri 16:41, 30 August 2005 (UTC)

Alfred, this is truly just the sort of input we needed! I agree with what you have said totally. I was a bit reluctant to name the central transistors as 'common base' as I dont see a low impedance at their bases. Perhaps you could elucidate on this minor problem?. However, on all other aspects, you have confirmed my thoughts. Thank you! PS User:Omegatron has kindly offered to label the transistors for us. Light current 16:56, 30 August 2005 (UTC)


 * I agree with you that technically, the base of a common base amplifier should be connected to 'common' so that there is no signal voltage present at the base. But, if you are going to be picky, a common emitter amplifier should not have an emitter resistor (or at least it should be bypassed at signal frequencies with a capacitor).  The transistors are common base amplifiers because the emitter is driven by an input transistor and the output is taken from the collector.  Further, the bases of these transistors (ideally) do not have any differential signal present so that they actually are 'common' for differential inputs.  Come to think of it, these transistors are common base amps for differential input signals but are actually difference amps for common mode input signals.  Think about that a little and see if you agree.  Alfred Centauri 17:21, 30 August 2005 (UTC)

Nice one Alfred ! I'll have to think about that. Light current 17:25, 30 August 2005 (UTC) Yes, I suppose if the inputs to the common base transistors are diff'l, then it doesnt matter about their bases being bypassed. How very clever! If the bases were to be bypassed to gnd tho', would that upset the common mode rejection? Light current 20:25, 30 August 2005 (UTC)

Not too sure about your last statement regarding CM response. As far as I can see any CM signals will try to turn on both transistors but fail to do so because there is nowhere for the extra base current to go. In this case, does the voltage impressed at the emitters just appear at the collectors, unamplified, or what? I dunno! Light current 20:46, 30 August 2005 (UTC) Seems that Q3,Q4 (the common base pair) are actually level shifters according to this.. Of course, how stupid of me not to realise this before. So it seems my previous steament 'the voltage impressed at the emitters just appears at the collectors' was correct after all ( but nor necessarily due to the correct thinking) Light current 21:17, 30 August 2005 (UTC)
 * They are more than level shifters, my friend! I perused the link and have found several major mistatements.

Yes there do seem to be some major bloomers here!Light current 22:51, 30 August 2005 (UTC)
 * (1) Q11 delivers current to the input stage - wrong! (Agreed LC)
 * (2) Q10 delivers Q11 current to Q3, Q4 bases - wrong! (Agreed LC)
 * (3) Q3, Q4 provide (just) DC level shifting - wrong! (Agreed LC)
 * (4) Q5,Q6,Q7, (etc.) provide gain - wrong! (Agreed - I think! LC)


 * But by all means, don't take my word for it. SPICE the (input) circuit.  Look at the AC voltage gain between the emitter and collector of Q4 - my bet is that it is WAY more than 1.  Look at it this way, if the emitter voltage of Q4 just appears at the collector of Q4, this entire stage has a gain < 1!  I'm likely to offend someone with the following statement, but here goes.  It is obvious to me that the presentation you linked to was prepared by an application engineer and not a design engineer.

Agreed -- and I'm not offended anyway! Light current
 * Regarding my previous statement: when a common mode input signal is present, the emitters and bases of Q3 and Q4 are both driven.  Thus, they act as difference amps.  The base drive signal comes from the collector of Q9.  If the common mode input signal is positive, the emitters of Q3, Q4 go more positive and the collector of Q9 will go more positive.  Thus, the bases of Q3 and Q4 also go more positive.  In other words, Q3 and Q4 ignore common mode input signals.  On the other hand, with differential input signals, the sum of the collector currents of the input transistors is constant.  This means that the collector voltage at Q9 is constant and thus, the base voltages of Q3 and Q4 are constant.  That is, the bases of Q3 and Q4 are at AC common. Alfred Centauri 22:14, 30 August 2005 (UTC)

Common mode signals
Large common mode input (ie same signal on both i/ps) would tend to increase the current drawn from the input of the upper left current mirror(Q8 & Q9). The output of the current mirror would then rise, tending to turn off the central transistor pair. Differential mode signals do not activate the current source in this way because the sum of the currents down each half of the stage is constant.
 * I agree. For the same reason that the DC bias is stabilized, the common-mode current is reduced. Alfred Centauri 17:01, 30 August 2005 (UTC)

Offset Null
If a pot is palced between the offset terminals, with the wiper going to V-, this can be be made to alter the current balance between both halves of the stage causing the output voltage of the stage to deviate from zero.(therby correcting for any offset) Do these explanations make sense to anyone? Light current 16:35, 29 August 2005 (UTC)
 * That's correct. Let the total resistance of the pot equal R and the position of the wiper equal x where 0 <= x <= 1.  With the pot connected as described, you have put a resistance of xR in parallel with one of the emitter resistors and a resistance of (1-x)R in parallel with the other.  If x <> 0.5, the quiescent current in each half of the mirror will no longer by equal.  As you said, this can be used to cancel the offset voltage. Alfred Centauri 17:09, 30 August 2005 (UTC)

Open Loop Output Impedance (resistance) of Op-Amp
Where do the figures of '1 ohm' and '1k ohm' come from?? Light current 23:28, 27 August 2005 (UTC)

It seems pretty obvious to me from the diagram that the output impedance without overall feedback has to be at least 25 ohms (for +ve outputs) and at least 50 ohms (for -ve outputs). Any comments? Light current 12:14, 28 August 2005 (UTC)
 * From a small signal perspective, the upper and lower emitter circuits are in parallel. Still, the output impedance must be higher than 16.7 ohms. Alfred Centauri 20:21, 30 August 2005 (UTC)

Im not too sure about being able to consider upper and lower transistors being in //, Alfred. If the o/p stage were class A, I would agree with you. But as it is class AB there wont be much standing current. The transistor that is only passing standing current will have a higher resistance than the other, wont it? So we have a different output resistance depending on the polarity of the output?. (open loop only) Light current 22:26, 30 August 2005 (UTC)